# 超急~四題物理題目求解釋!!

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1a) 1200 rev is equal to 1200 x 2π = 2400π rad.

Hence the angular speed is ω = 2400π/60 = 40π = 126 rad/s

b) Tangential speed = rω = 0.03 x 40π = 3.77 m/s

c) Centripetal acc. = rω2 = 0.08 x (40π)2 = 1.26 km/s2

d) Tangential speed on the rim = 0.08 x 40π = 10.05 m/s

Hence total distance moved in 2 s = 10.05 x 2 = 20.1 m

2a) 0.8 x 30 = 24 Nm

b) Linear acc. = 0.8/0.75 = 1.07 m/s2

So angular acc. = 1.07/30 = 0.0356 rad/s2

c) 0.0356 x 30 = 1.07 m/s2

3a) Angular acc. of the wheel = 10/6 = 5/3 rad/s2

So moment of inertia = 36/(5/3) = 21.6 kg m2

b) Angular decelaration of the wheel = 10/60 = 1/6 rad/s2

Hence torque dur to friction = 21.6 x 1/6 = 3.6 N m

c) Angular displacement in the period of acc. is:

(1/2) x (5/3) x 102 = 250/3 rad

Angular displacement in the period of dec. is:

10 x 60 - (1/2) x (1/6) x 602 = 300 rad

So total angular displacement = 300 + 250/3 = 1150/3 rad

With no. of rev = 1150/(3 x 2π) = 52.5

4) Torque produced = 50 x 1.5 = 75 N m

Mass of the round = 80 kg, with moment of inertia = 0.5 x 80 x 1.52 = 90 kg m2.

So angular acc. = 75/90 = 5/6 rad/s2

After 3 s, angular velocity = 5/6 x 3 = 2.5 rad/s

So k.e. = 0.5 x 90 x 2.52 = 281.25 J

Source(s): 原創答案
• 應付期末考吧= =+